A fair coin is tossed $n$ times. Let $X$ denote the number of times head occurs. If $P(X = 4)$, $P(X = 5)$ and $P(X = 6)$ are in arithmetic progression, then the value of $n$ can be:
Step-by-Step Solution
Key Concept: The arithmetic progression condition for binomial coefficients converts to an algebraic equation by expanding the factorials and simplifying.
Given that $^nC_4$, $^nC_5$, and $^nC_6$ are in arithmetic progression, we use $2 · ^nC_5 = ^nC_4 + ^nC_6$. Expanding binomial coefficients and simplifying: $\frac{2(n-4)}{5} = 1 + \frac{(n-4)(n-5)}{5·6}$ leads to $n^2 - 21n + 98 = 0$, giving $(n-7)(n-14) = 0$. Thus $n = 7$ or $n = 14$.
Correct Answer: 1,4